Question

If ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 1}&1 \\
  { - 15}&6&{ - 5} \\
  5&{ - 2}&2
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  1&2&{ - 2} \\
  { - 1}&3&0 \\
  0&{ - 2}&1
\end{array}} \right]$, find \[{(AB)^{ - 1}}\].

Answer 1

Hint: First we will find the determinant of the matrix $\left( {{A^{ - 1}}} \right)$ then we will find the adjoint of the same matrix, now using these we will use the formula for the inverse of a matrix hence will get a matrix $\left( A \right)$.
Now, using this and the given matrix we will find the required answer.

Complete step-by-step answer:
Given data: ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  3&{ - 1}&1 \\
  { - 15}&6&{ - 5} \\
  5&{ - 2}&2
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  1&2&{ - 2} \\
  { - 1}&3&0 \\
  0&{ - 2}&1
\end{array}} \right]$
Here we will the formula ${(X)^{ - 1}}$ to find the matrix of that is given by
${(X)^{ - 1}} = \dfrac{1}{{\left| X \right|}}adj(X)$
We also know that ${\left( {{A^{ - 1}}} \right)^{ - 1}} = A$
For $\left( {{A^{ - 1}}} \right)$, we can say that $A = \dfrac{1}{{\left| {{A^{ - 1}}} \right|}}adj\left( {{A^{ - 1}}} \right)$
Now for the determinant of $\left( {{A^{ - 1}}} \right)$ matrix that is
$\left| {{A^{ - 1}}} \right| = \left| {\begin{array}{*{20}{c}}
  3&{ - 1}&1 \\
  { - 15}&6&{ - 5} \\
  5&{ - 2}&2
\end{array}} \right|$
$ = 3\left( {12 - 10} \right) - \left( { - 1} \right)\left( { - 30 + 25} \right) + 1\left( {30 - 30} \right)$
On simplifying the brackets
$ = 3\left( 2 \right) + \left( { - 5} \right)$

$ = 1$
Now $adj\left( {{A^{ - 1}}} \right)$ is the transpose of the cofactor matrix of $\left( {{A^{ - 1}}} \right)$.
A cofactor is the determinant of the matrix excluding the column and the row of that particular element
multiplied with a factor of ${\left( { - 1} \right)^{i + j}}$, where i and j are the position column and row of the element respectively.
Now the cofactors are
${A^{ - 1}}_{11} = {( - 1)^2}\left| {\begin{array}{*{20}{c}}
  6&{ - 5} \\
  { - 2}&2
\end{array}} \right|$
On simplification we get,
$ = \left( {12 + 10} \right)$
$ = 22$
\[{A^{ - 1}}_{12} = {( - 1)^3}\left| {\begin{array}{*{20}{c}}
  { - 15}&{ - 5} \\
  5&2
\end{array}} \right|\]
On simplification we get,
$ = ( - 1)\left( { - 30 + 10} \right)$
$ = 20$
${A^{ - 1}}_{13} = {( - 1)^4}\left| {\begin{array}{*{20}{c}}
  { - 15}&6 \\
  5&{ - 2}
\end{array}} \right|$
On simplification we get,
$ = \left( {30 - 30} \right)$
$ = 0$
${A^{ - 1}}_{21} = {( - 1)^3}\left| {\begin{array}{*{20}{c}}
  { - 1}&1 \\
  { - 2}&2
\end{array}} \right|$
On simplification we get,
$ = ( - 1)\left( { - 2 + 2} \right)$
$ = 0$
${A^{ - 1}}_{22} = {( - 1)^4}\left| {\begin{array}{*{20}{c}}
  3&1 \\
  5&2
\end{array}} \right|$
On simplification we get,
$ = \left( {6 - 5} \right)$
$ = 1$
${A^{ - 1}}_{23} = {( - 1)^5}\left| {\begin{array}{*{20}{c}}
  3&{ - 1} \\
  5&{ - 2}
\end{array}} \right|$
On simplification we get,
$ = ( - 1)\left( { - 6 + 5} \right)$
$ = 1$
${A^{ - 1}}_{31} = {( - 1)^4}\left| {\begin{array}{*{20}{c}}
  { - 1}&1 \\
  6&{ - 5}
\end{array}} \right|$
On simplification we get,
$ = \left( {5 - 6} \right)$
$ = - 1$
${A^{ - 1}}_{32} = {( - 1)^5}\left| {\begin{array}{*{20}{c}}
  3&1 \\
  { - 15}&{ - 5}
\end{array}} \right|$
On simplification we get,
$ = ( - 1)\left( { - 15 + 15} \right)$
$ = 0$
${A^{ - 1}}_{33} = {( - 1)^6}\left| {\begin{array}{*{20}{c}}
  3&{ - 1} \\
  { - 15}&6
\end{array}} \right|$
On simplification we get,
$ = \left( {18 + 15} \right)$
$ = 33$
Therefore, transpose of \[adj\left( {{A^{ - 1}}} \right) = \left[ {\begin{array}{*{20}{c}}
  {22}&{20}&0 \\
  0&1&1 \\
  { - 1}&0&{33}
\end{array}} \right]\]
Therefore, \[adj\left( {{A^{ - 1}}} \right) = \left[ {\begin{array}{*{20}{c}}
  {22}&0&{ - 1} \\
  {20}&1&0 \\
  0&1&{33}
\end{array}} \right]\]
Now, \[A = \dfrac{1}{{\left| {{A^{ - 1}}} \right|}}adj\left( {{A^{ - 1}}} \right)\]
Substituting the value if \[\left| {{A^{ - 1}}} \right|\]and \[adj\left( {{A^{ - 1}}} \right)\]
\[ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
  {22}&0&{ - 1} \\
  {20}&1&0 \\
  0&1&{33}
\end{array}} \right]\]
Now we have to find \[{(AB)^{ - 1}}\]
$ \Rightarrow {(AB)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {22}&0&{ - 1} \\
  {20}&1&0 \\
  0&1&{33}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2&{ - 2} \\
  { - 1}&3&0 \\
  0&{ - 2}&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  {22}&{44 + 2}&{ - 44 - 1} \\
  {20 - 1}&{40 + 3}&{ - 40} \\
  { - 1}&{3 - 66}&{33}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
On simplifying the first matrix, we get,
$ = \left[ {\begin{array}{*{20}{c}}
  {22}&{46}&{ - 45} \\
  {19}&{43}&{ - 40} \\
  { - 1}&{ - 63}&{33}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
On subtracting we get,
$ = \left[ {\begin{array}{*{20}{c}}
  {21}&{46}&{ - 45} \\
  {19}&{42}&{ - 40} \\
  { - 1}&{ - 63}&{32}
\end{array}} \right]$
Therefore the required answer is $\left[ {\begin{array}{*{20}{c}}
  {21}&{46}&{ - 45} \\
  {19}&{42}&{ - 40} \\
  { - 1}&{ - 63}&{32}
\end{array}} \right]$

Note: While finding the adjoint of a matrix most of the students forget to take the transpose of the matrix formed of cofactors, which results in the incorrect matrix for the inverse.
So remember that adjoint of any matrix is the transpose of cofactors.